Practice Number System Aptitude Questions:

Overview:

 Questions and Answers Type: MCQ (Multiple Choice Questions). Main Topic: Quantitative Aptitude. Quantitative Aptitude Sub-topic: Number System Aptitude Questions and Answers. Number of Questions: 10 Questions with Solutions.

1. Find the LCM of 21, 28, and 32?

1. 572
2. 672
3. 678
4. 770

Solution:

Step(1): Factorize the numbers into their prime factors.$$21 = 3 \times 7 = 3^{1} \times 7^{1}$$ $$28 = 2 \times 2 \times 7 = 2^{2} \times 7^{1}$$ $$32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^{5}$$ Step(2): Collect all the distinct factors with their maximum available power.$$= 2^{5}, \ 3^{1}, \ 7^{1}$$ Step(3): Multiply the collected factors.$$= 2^{5} \times 3^{1} \times 7^{1}$$ $$= 32 \times 3 \times 7 = 672$$

1. Find the HCF of $$\frac{6}{7}$$ and $$\frac{7}{8}$$?

1. $$\frac{1}{56}$$
2. $$\frac{1}{58}$$
3. $$\frac{1}{62}$$
4. $$\frac{1}{66}$$

Answer: (a) $$\frac{1}{56}$$

Solution: $$HCF \ of \ fractions = \frac{HCF \ of \ Numerators}{LCM \ of \ Denominators}$$
$$HCF = \frac{HCF \ of \ (6,7)}{LCM \ of \ (7,8)}$$ $$= \frac{1}{56}$$

1. Find the HCF of 15, 25, 45?

1. 2
2. 3
3. 4
4. 5

Solution: Step(1): Factorize the numbers into their prime factors.$$15 = 3 \times 5 = 3^{1} \times 5^{1}$$ $$25 = 5 \times 5 = 5^{2}$$ $$45 = 3 \times 3 \times 5 = 3^{2} \times 5^{1}$$ Step(2): Collect all the common factors with their minimum available power.$$= 5^{1} = 5$$ Here, 5 is the highest positive number that can divide 15, 25, and 45 exactly.

1. Which one of the following is the LCM of $$\frac{15}{4}$$, $$\frac{18}{5}$$, and $$\frac{20}{7}$$?

1. 120
2. 150
3. 180
4. 220

Solution: $$LCM \ of \ fractions = \frac{LCM \ of \ Numerators}{HCF \ of \ Denominators}$$ $$LCM = \frac{LCM \ of \ (15, 18, 20)}{HCF \ of \ (4, 5, 7)}$$ $$= \frac{180}{1} = 180$$

1. Find the HCF of 120, 150, 180?

1. 10
2. 20
3. 30
4. 60

Solution: Step(1): Factorize the numbers into their prime factors.$$120 = 2 \times 2 \times 2 \times 3 \times 5$$ $$= 2^{3} \times 3^{1} \times 5^{1}$$ $$150 = 2 \times 3 \times 5 \times 5 = 2^{1} \times 3^{1} \times 5^{2}$$ $$180 = 2 \times 2 \times 3 \times 3 \times 5$$ $$= 2^{2} \times 3^{2} \times 5^{1}$$ Step(2): Collect all the common factors with their minimum available power.$$= 2^{1}, \ 3^{1}, \ and \ 5^{1}$$ Step(3): Multiply the collected factors.$$= 2^{1} \times 3^{1} \times 5^{1}$$ $$= 2 \times 3 \times 5 = 30$$ Here, 30 is the highest positive number that can divide 120, 150, and 180 exactly.

1. Find the LCM of $$\frac{65}{20}$$, $$\frac{75}{35}$$, and $$\frac{80}{9}$$?

1. 15600
2. 15750
3. 15820
4. 15950

Solution: $$LCM \ of \ fractions = \frac{LCM \ of \ Numerators}{HCF \ of \ Denominators}$$ $$LCM = \frac{LCM \ of \ (65, 75, 80)}{HCF \ of \ (20, 35, 9)}$$ $$= \frac{15600}{1} = 15600$$

1. Which of the following is the HCF of 49, 64, and 99?

1. 1
2. 2
3. 3
4. 4

Solution: Step(1): Factorize the numbers into their prime factors.$$49 = 7 \times 7 = 7^{2}$$ $$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{6}$$ $$99 = 3 \times 3 \times 11 = 3^{2} \times 11^{1}$$ There is no common number among the factors, so the HCF must be 1.

1. Which one of the following is the LCM of 7, 11, 13, and 17?

1. 1
2. 17017
3. 7
4. 65

Solution:

Step(1): Factorize the numbers into their prime factors.$$7 = 7^{1}$$ $$11 = 11^{1}$$ $$13 = 13^{1}$$ $$17 = 17^{1}$$Step(2): Collect all the distinct factors with their maximum available power.$$= 7^{1}, \ 11^{1}, \ 13^{1}, \ and \ 17^{1}$$ Step(3): Multiply the collected factors.$$= 7 \times 11 \times 13 \times 17 = 17017$$

1. Find the LCM of $$\frac{10}{9}$$, $$\frac{12}{11}$$, and 25?

1. 2
2. 150
3. 300
4. 600

Solution: $$LCM \ of \ fractions = \frac{LCM \ of \ Numerators}{HCF \ of \ Denominators}$$ $$LCM = \frac{LCM \ of \ (10, 12, 25)}{HCF \ of \ (9, 11, 1)}$$ $$= \frac{300}{1} = 300$$

1. Find the HCF of 18, 24, and $$\frac{6}{5}$$?

1. $$\frac{7}{5}$$
2. $$\frac{4}{5}$$
3. $$\frac{8}{5}$$
4. $$\frac{6}{5}$$

Answer: (d) $$\frac{6}{5}$$

Solution: $$HCF \ of \ fractions = \frac{HCF \ of \ Numerators}{LCM \ of \ Denominators}$$
$$HCF = \frac{HCF \ of \ (18, 24, 6)}{LCM \ of \ (1, 1, 5)}$$ $$= \frac{6}{5}$$